Negative Conductors and
Network Planarity
Konrad Schrøder
26 February 1993
Section 1: Introduction
In this paper we consider conductor networks which possibly contain both positive and negative conductors.
A conductor is a two-sided object which obeys Ohm’s law: if potentials V1 and V2 are applied respectively
to each end of the conductor, then the current flowing through the conductor is given by
I = (V1 − V2 )γ
where γ is an intrinsic property of the conductor, known as its conductance. When we speak, therefore, of
“positive” and “negative” conductors, we mean conductors whose conductances are positive and negative,
respectively.
A network of conductors is any number of conductors, joined at the ends; such joinings are known as
nodes. If all conductances are positive, this can be represented physically by a resistor network, or more
generally by a network of pipes or channels. In the networks that we consider, certain nodes form the
boundary of the network, and others the interior; we restrict our ability to impose potentials or currents to
the boundary only.
Because we have so restricted ourselves, Kirkhoff’s current law will hold at each of the interior nodes; in
the case where all conductances are positive, this implies a unique set of interior potentials due to imposed
boundary potentials (the Dirichlet problem) and a unique (up to a constant) set of potentials due to imposed
boundary currents (the Neumann problem). This uniqueness is essential to the construction of certain useful
tools, such as the Dirichlet-to-Neumann map.
If some of the conductors are negative, the situation is more complicated: in the case of electrical
networks, a negative conductor would represent a non-dissipative element, such as a generator; similarly it
would represent a pump in the case of a pipe network. Uniqueness of the Dirichlet and Neumann problems
is not automatic in these cases: for example, two resistors of resistance 1 and −1 placed in series will
make a effectively resistance-free object; current may flow when both ends are grounded, meaning that the
interior potential is something other than zero. Similarly, the same resistors placed in parallel1 will form a
short circuit; and current might flow through the network even though both ends are insulated. Other such
combinations, however, do not suffer from this problem: replace −1 by − 21 in each case and no badly-behaved
object is created.
We are interested primarily with non-singular networks, that is, networks that do not exhibit this bad
behaviour, and section 3 concentrates on this topic; the first few sections on singular networks are provided
for background.
Section 2: Singular Networks
1
Technically, such multiply-connected graphs will not be allowed in our discussion; however dividing one
of the conductors into two will yield a legal network with the same property; e.g., resistances 12 and 12 in
series, in parallel with resistance −1.
1
Section 2.1: general definitions; the Neumann problem
We begin with the Neumann problem, since it is easier to analyze than the Dirichlet problem. In particular,
imposing zero current on the boundary nodes is equivalent to requiring that Kirkhoff’s current law hold
there; so the distinction between boundary and interior nodes is only nominal when considering solutions to
the homogeneous Neumann problem.
Definition 2.1. A network consists of a graph Ω = (N, E) along with a conductivity function γ defined on
the set E of edges, such that γ 6= 0. The set N of nodes is divided into two subsets: the set Nb of boundary
nodes and the set Ni of interior nodes.
Definition 2.2. A function f defined on the nodes of a network is said to be γ-harmonic if at every interior
node i, it is true that Kirkhoff’s current law holds, i.e.,
X
X
f (j)γij = f (i)
γij
j
j
Definition 2.3. For any function f defined on the nodes of a network, the boundary current at a boundary
node i is given by
X
(f (i) − f (j))γij
j
The statement of the Neumann problem on conductor networks is this: given a vector ψ of boundary
currents, is there a γ-harmonic function f defined on the nodes of Γ such that ψ gives the boundary current
for f ? If the conductances γ are known to be all positive, then the existence and uniqueness—up to a
constant—of f are guaranteed. For conductances both positive and negative, however, neither of these
things is guaranteed, since current can flow in circles.
If there are no “circles”, however, for current to flow around, then we should have a well-behaved
network. We look first, therefore, at networks whose graphs are trees; and then generalize to graphs that
contain circuits.
Lemma 2.1. Suppose that u is the solution to the homogeneous Neumann problem on a network Γ, and
that i is a node of valence 1 in Γ. Let Γ0 be the network obtained from Γ by deleting i and the edge ij,
where j is the sole neighbor of i. Then u|Γ0 is a solution to the homogeneous Neumann problem on Γ0 ; and
u(i) = u(j).
Proof. Kirkhoff’s law at i states that u(i) = u(j); since this is so,
X
X
(u(j) − u(k))γjk =
(u(j) − u(k))γjk
k
k6=i
Since u satisfied Kirkhoff’s law at j, then u|Γ0 satisfies Kirkhoff’s law at j.
Theorem 2.1. Given a network Γ, and a subnetwork T whose underlying graph is a tree. Further suppose
that there exists some node i, an endpoint of T , such that the only way to reach any node of T from a node
not in T is to pass through i. Denote by S the network Γ \ T , where both i ∈ S and i ∈ T . Then if u
is a solution to the homogeneous Neumann problem on Γ, u|S is a solution to the homogeneous Neumann
problem on S; and furthermore, for any node j ∈ T , u(j) = u(i).
Proof. Apply lemma 2.1 to the endpoints of T repeatedly until T has been entirely removed from
the network, except for node i.
Corollary 2.1. Suppose Γ is a network whose underlying graph is a tree. Then the family of solutions of
the homogeneous Neumann problem is one-dimensional.
Theorem 2.2. Given any network Γ. The family of solutions to the homogeneous Neumann problem on Γ
is at most (nc + 1)-dimensional, where nc is the number of basis circuits in the graph underlying Γ.
Proof. Let Γ0 be Γ restricted to a tree spanning its graph. By corollary 2.1, the only solutions to
the homogeneous Neumann problem on Γ0 are constants. For every edge e ∈ (Γ \ Γ0 ), we specify a
2
current ie . The current ie must flow through the circuit closed by the addition of e to Γ0 ; if it does
not, then some part of the current must flow across another edge e2 not in Γ0 ; but then the current
on e2 would not be what we specified.
There are nc such edges in Γ; thus the family of solutions to the homogeneous Neumann
problem on Γ is at most nc + 1-dimensional.
Theorem 2.3. Suppose that Γ is a network, none of whose circuits is singular, i.e. such that the sum of the
inverse of the conductances along any circuit is not 0. Then the only solutions to the homogeneous Neumann
problem on Γ are constants.
Proof. Consider a spanning tree Γ0 as in the proof of theorem 2.2; and let AB be an edge of Γ not
in Γ0 . Applying Ohm’s law repeatedly to the edges of the circuit C closed by AB gives
X
u(A) − u(B) = −
e∈C, e6=AB
iAB
γe
where iAB is the current flow from A to B. We can also apply Ohm’s law directly to AB:
u(A) − u(B) =
iAB
γAB
In order for these to agree, we must have
X iAB
=0
γe
e∈C
Since this condition does not hold for any circuit C, the current iAB is equal to zero. Since this
is so, Kirkhoff’s current law at A and B will remain unchanged if the edge AB is removed. Thus,
the only solutions to the homogeneous Neumann problem are those solving the same problem on
Γ0 ; i.e., constants.
Section 2.2: the Dirichlet problem.
Now we turn to the Dirichlet problem. Here, the difference between interior and boundary nodes does not
simply go away; also, we must consider the possibility of singular elements; if a network with 2 boundary
nodes A and B can have a potential difference but no current flow between A and B, then it is disconnected
and thus illegal; but such a network with current flowing and no potential difference between A and B is
not in any way illegal.
Definition 2.4. The interior of a network Γ = (N, E, γ), denoted by Γ◦ is the network (N, E 0 , γ|E 0 ) obtained
from Γ by deleting all edges that join two boundary nodes.
Definition 2.5. The skeleton of a network Γ is the network obtained by deleting all boundary nodes and
all edges that abut boundary nodes.
Definition 2.6. A network is said to be a tree network if the graph underlying its skeleton is a tree.
3
Definition 2.7. A tree network is said to be a linear tree network if its skeleton consists of a single chain.
Now that our terms are well-defined, we begin our discussion of the Dirichlet problem on conductor
networks. We first consider linear tree graphs, where the question of the uniqueness of solutions to the
Dirichlet problem is relatively simple; we can then use these linear trees as building blocks to build up larger
and more complicated graphs.
Lemma 2.2. If u is a solution to the homogeneous Dirichlet problem on a network Γ, then u is also a
solution of the same problem on Γ◦ .
Proof. Suppose that u is a solution to the homogeneous Dirichlet problem on a network Γ. Since
the potential at every boundary node is zero, Ohm’s law applied to any boundary-to-boundary
conductor gives 0 = 0γ; and we can choose γ arbitrarily to satisfy this equation. In particular, we
can choose γ = 0.
Lemma 2.3. If u is a solution to the homogeneous Dirichlet problem on a network Γ, and Γ0 is the network
formed by removing all interior nodes of valence 1, then u|Γ0 is a solution to the homogeneous Dirichlet
problem on Γ0 .
Proof. Suppose i is an interior node of valence 1. Since Kirkhoff’s law holds at i, no current ever
flows into or out of i. We can therefore delete the edge joining i to its neighbor without disturbing
the functioning of Kirkhoff’s law at the neighbor.
By these lemmas, we can ignore boundary-to-boundary connections when considering whether the
Dirichlet problem has a unique solution on a network; and we can ignore sections of the graph that are
“almost entirely insulated”. The following discussion assumes that the networks in question do not have
either of these features.
Theorem 2.4. The family of solutions to the homogeneous Dirichlet problem on a linear tree network Γ is
at most one-dimensional.
Proof. Number the interior nodes 1, 2, . . ., ni , so that nodes k and k + 1 are connected for any k
strictly between 0 and ni . Suppose that we knew the potential at node k < ni and at node k − 1.
Since the network is a linear tree network, node k is connected to at most two other interior nodes,
k − 1 and k + 1. Knowing the potentials at all neighbors of node k but node k + 1, Kirkhoff’s law
will determine the potential at node k + 1.
Suppose that we choose a potential x at node 1. By induction, we have determined the
potentials at nodes 2, 3, . . ., ni . Thus the possible solutions to the Dirichlet problem on a linear
tree network form a at most a one-parameter family.
Corollary 2.2. Suppose that the potential at all boundary nodes of a linear tree network is zero, and that
the potential at one end of its skeleton is known. Then the potential function u satisfying Kirkhoff’s law at
every point of the interior is unique, if it exists.
Definition 2.8. A network is said to be D-singular if any nontrivial solution to the homogeneous Dirichlet
problem exists on that network.
Theorem 2.3. Suppose that a linear tree network is D-singular. Then any non-trivial solution u to the
homogeneous Dirichlet problem on that network must have non-zero current flow at boundary nodes that
are connected to nodes at the ends of the skeleton.
Proof. Suppose that some solution had zero current flow at some boundary node B, which is
connected to an interior node I, which in turn is one endpoint of the skeleton of the network. Since
the network is a linear tree network, B is connected to no other interior node. Ohm’s law applied
to the conductor IB implies that u(I) = u(B) = 0. Again applying Ohm’s law to any conductors
that join I to boundary nodes, we find that all currents between I and boundary nodes are zero.
Now number the interior nodes I0 , I1 , . . . , In ; suppose that node Ik−1 has potential zero, and
that no current flows into Ik−1 from another interior node (possibly excepting Ik ). Applying
Ohm’s law to all connections joining Ik−1 to boundary nodes, we see that no current flows into
Ik−1 from the boundary; applying Kirkhoff’s law, we see further that no current flows out of Ik−1 ,
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and in particular no current flows between Ik−1 and Ik . Using Ohm’s law once again, we see that
u(Ik ) = u(Ik−1 ) = 0.
Thus we know by induction that the potential on every interior node is zero. But then u is
not a non-trivial solution, which is a contradiction.
Theorem 2.4. The family of solutions to the homogeneous Dirichlet problem on a tree network is at most
(nl − 1)-dimensional, where nl denotes the number of leaves of the graph underlying its skeleton.
Proof. Consider the network as a collection of subnetworks, each of which are linear tree networks.
These subnetworks intersect at branch points, namely those nodes of the skeleton which have valence
greater than two. Each subnetwork component touches either one branch point or two; if it touches
only one, let the node at the other end of its skeleton be called a leaf point. Thus the nodes of the
skeleton are divided into three classes: leaf points, which have valence 1; branch points, which have
valence > 2; and other points, which are not important to our discussion. Denote by bi the valence
of the branch point i; and by nl the number of leaf points.
Consider a branch point i, with known potential ui and interior neighbor j whose potential is
known. We can arbitrarily choose potentials at all of the other interior neighbors but one; and at
the final interior neighbor Kirkhoff’s law will uniquely determine the potential.
Consider the component subnetworks abutting node i, apart from that containing node j. By
Corollary XXX, the potential in these components, up to and including their other branch points
(if any), is determined.
Now, we can traverse the tree recursively, considering any leaf point as the tree’s root; our first
choice of potential is at the root, which determines potential at the first branch point; and at every
other branch point i we can make bi − 2 choices of P
potential.
In this way we have constructed an at most ( i (bi − 2) + 1)-dimensional family of solutions
to the homogeneous Dirichlet problem on tree networks. Since for any tree
X
(bi − 2) = nl − 2,
i
this number can also be written as nl − 1.
In fact, it is possible to realize this bound for any given shape of tree simply by making all components
singular; in this case the potential at every branch point is 0; and we can choose nl −1 independent parameters
that will result in a solution to the homogeneous Dirichlet problem.
Theorem 2.5. The family of solutions to the homogeneous Dirichlet problem on a network Γ is at most
(nl + nc − 1)-dimensional, where nl is the number of leaf points in a tree network spanning Γ, and nc is the
number of basis circuits in the graph underlying the skeleton of Γ.
Proof. Consider the skeleton of Γ. We can form a new network Γ0 by deleting enough linear tree
subnetworks from Γ that the skeleton of Γ0 is a tree; we can then apply theorem 2.4 to Γ0 . Now
suppose that we added one of the linear trees, T , that is in Γ but not in Γ0 , thus forming a circuit in
the new skeleton. We may be able to specify the volatge at some point in T ; if we can, this gives us
one more free parameter. In adding Γ \ Γ0 , we have obtained at most nc parameters; therefore, the
family of solutions to the nomogeneous Dirichlet problem on Γ is at most (nl + nc − 1)-dimensional.
In some sense, this result is not as good as theorem 2.4, since for not all shapes of graph can the
bound be realized; for example, networks whose skeleton consists of a single circuit consistently have only a
1-dimensional family of solutions, whereas the count in theorem 2.5 would have the dimension be 2+1−1 = 2.
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3. Relation to Planarity
3.1. Circular-planar graphs and the Alternating Principle
When all conductors are constrained to be positive, [3] gives a characterization of the set Ω of circular-planar
graphs on the basis of signs of certain subdeterminants of the response matrices.
Definition 3.1. A Circular Planar Graph is a graph embedded in a disc D in the plane so that the boundary
nodes lie on the circle C which bounds D, and the reast of Γ is in the interior of D.
Definition 3.2. Let Γ be a circular planar graph, with boundary nodes v1 , . . . , vn . in clockwise order around
C. A pair of sequences of boundary nodes (P ; Q) = (p1 , . . . , pk ; q1 , . . . , qk ) such that the entire sequence
(p1 , . . . , pk , q1 , . . . , qk ) is in circular order is called a Circular Pair.
Definition 3.3. Let Γ be a circular planar graph, with boundary ndoes v1 , . . . , vn , and response matrix Λ;
and let (P ; Q) be a circular pair of boundary nodes. The submatrix obtained by taking the entries of Λ that
are in rows p1 , . . . , pk and in columns q1 , . . . , qk will be called a circular submatrix of Λ, and will be denoted
Λ(P ; Q). det Λ(P ; Q) will be known as a k × k circular subdeterminant of Λ.
Theorem 3.1. (Curtis and Morrow) Let Γ be a circular planar network, and let Λ be the corresponding
k
response matrix. Then for any circular pair (P ; Q), −1d 2 e det Λ(P ; Q) > 0.
Theorem 3.2. (Curtis and Morrow) Let Λ be an n × n matrix in Ω+ , and also suppose that for any
k
circular pair (P ; Q), −1d 2 e det Λ(P ; Q) > 0. Then there exists some circular planar network Γ having Λ as
its response matrix.
Proof. The proof for theorems 3.1 and 3.2 is provided in [3].
It is clear, then, that the response matrices of graphs containing negative conductors cannot lie in Ω+ :
Corollary 3.1. Let (Γ, γ) be a critical, recoverable, circular planar network containing at least one negative
conductor, and let Λ be the corresponding response matrix. Then Λ 6∈ Ω+ .
Proof. Suppose otherwise: Λ ∈ Ω+ . Since Γ is a critical graph, by theorem 3.2, there exist positive
conductances γ 0 on Γ such that Λ is the response matrix for (Γ, γ 0 ). However, since Γ is recoverable,
there is exactly one set of conductances on Γ that will yield the response matrix Λ. So γ = γ 0 . But
this cannot be, since γ 0 is all positive, and γ is known to have at least one negative value.
6
3.2. Equivalence between mixed-sign and non-planar networks
However, it is possible to find mixed-sign conductor networks whose response matrix is equivalent to that of
a non-planar graph. Consider, for example, the graph Σ6 :
Depending on the values of neighboring conductors, the conductor marked γ can be taken to have any of
a range of negative values, −∞ < γ < γ0 < 0, and the resulting response matrix Λ may still lie in K 6 ,
the space of 6-dimensional Kirkhoff matrices of positive conductance. To see this, we may remove γ and its
neighboring conductors, and replace them with a non-planar but electrically equivalent object:
For each of the objects N and P , we can construct a response matrix ΛN and ΛP respectively; since
these objects are supposed to be electrically equivalent, we then have ΛN = ΛP . In this case, we have
Σ
−ab(c + d + e)
1
−ab(c
+
d
+
e)
Σ

ΛP =

−acd
−bcd
(a + b + c)(c + d + e) − c
−ace
−bce


−acd
−ace
−bcd
−bce


Σ
−de(a + b + c)
−de(a + b + c)
Σ
Since the graph N is fully connected, we can simply read the conductances from the response matrix.
If all of the entries are to have the right sign when c < 0, then we must also have a + b + c < 0, c + d + e < 0,
and (a + b + c)(c + d + e) − c2 < 0, i.e., c < − (a+b)(d+e)
a+b+d+e . However, this last condition is redundant, since if
xy
x, y > 0, then x+y
< x + y; thus, the appropriate condition is
c + max{a + b, d + e} < 0
This method is easily generalizable to arbitrary graphs.
Theorem 3.3. Suppose that we are given a graph (Γ, γ), and a pair of nodes A and B in Γ such that A
and B are connected by a single conductor with conductance γAB < 0, but that all other conductances are
positive; and denote by N (x) the set of all conductors with x as one of their endpoints. Then the response
matrix Λγ ∈ K provided that
X
γpA < 0
(3.3.1)
γqA < 0
(3.3.2)
ΣA ΣB
ΣA + Σ B
(3.3.3)
p∈N (A)
X
q∈N (B)
and
γAB <
Proof. Let Γ0 be the set of nodes N (A) ∪ N (B), together with all edges joining those nodes to each
other, and consider the response matrix ΛΓ0 . Denote by ΣA the sum of all conductances at A, and
7
γ
γ
Σ
B
; if q ∈ N (A)
likewise define ΣB . If node p ∈ N (A), q ∈ N (A), p 6= q, then λΓ0 :p,q = − ΣApAΣBpB
−γ 2
γ
γ
AB
γ
pB AB
as well, λΓ0 :p,q = − ΣpA
. If γAB < 0, the off-diagonal entries will have mixed signs unless
2
A ΣB −γAB
(3.3.1) and (3.3.2) hold; and the sign of the entries in the upper-right and lower-left blocks will be
wrong unless (3.3.3) holds. However, if those conditions are satisfied, ΛΓ0 will be a viable response
matrix; since the rest of Γ has positive conductance, this means that ΛΓ is also a viable response
matrix, i.e., ΛΓ ∈ K.
Lemma 3.4. Suppose
P that M is a 4 × 4 matrix satisfying the following conditions: (1) mij < 0 for all
i, j, i 6= j; (2) mii = j6=i mij ; (3) mij = mji for all i, j. (I.e., M is a Kirkhoff matrix for a connected graph
of positive conductors.) Then there exists a set of conductances γ on Σ4 such that M = Λγ .
Proof. Let Σ4 be labelled as in the illustration. We can use a standard recovery technique to
attempt to recover the conductances γ:
x = −m14 /m13
(3.4.1)
γe = xm43 + m44
(3.4.2)
γd = m43 + m44 /x
(3.4.3)
γf = m23 + m24 /x
(3.4.4)
We can see that in order to solve for γd , γe , and γf , we must have certain conditions on M ;
namely, m13 6= 0, from (3.4.1), and m14 6= 0, from (3.4.3) and (3.4.4). Similarly, recovering the lefthand ladder of conductors, we find the conditions m42 6= 0 and m41 6= 0. However, since we know
already that mij 6= 0, all of these conditions are met, and we can solve uniquely for γa , γb , γd , γe , γf .
Solving now for γc , we place a potential of +1 at node 1, and 0 at all other boundary nodes; then,
w = m11 /γa
(3.4.5)
z = m14 /γe
(3.4.6)
γc = (m11 − m12 )/(x − y)
(3.4.5) implies that m14 (m12 +m11 ) 6= 0; since we already know that m14 6= 0, this simply means
that m12 + m11 6= 0. However, the only way that this could be false would be if m13 = m14 = 0,
which cannot be true. Similarly for the second condition: m14 (m43 +m44 ) 6= 0 is true, since m14 6= 0
and since M is a Kirkhoff matrix. To find γc , we need to know that x − y 6= 0. If x − y = 0, no
current could flow through the conductor c regardless of its conductance; similarly, no current could
flow through conductor f , regardless of its conductance. Therefore, m13 = −m14 , which cannot be
true.
Once we see that these conditions are satisfied, the fomulae give the conductances.
Conjecture 3.5. Suppose that M is a n × n Kirkhoff matrix, n > 3, such that mij 6= 0 for all i, j. Then
there exists a set of conductances γ on Σn such that M = Λγ .
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References
[1] Curtis, Edward B. and Morrow, James A., Determining the Resistors in a Network. SIAM J. Appl.
Math, Vol. 50, No. 3, pp. 918-930, June 1990.
[2] Curtis, Edward B. and Morrow, James A., The Dirichlet to Neumann Map for a Resistor Network. SIAM
J. Appl. Math, Vol. 51, No. 4, pp. 1011-1029, August 1991.
[3] Curtis, E. B., Ingerman, David, and Morrow, J. A. Circular Planar Graphs and Resistor Networks. [prepublication]
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